MZL's xor
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456 Accepted Submission(s): 322
Problem Description
Ai+
Aj)(
1≤i,j≤n)
The xor of an array B is defined as
B1 xor
B2...xor
Bn
Input
Multiple test cases, the first line contains an integer T(no more than 20), indicating the number of cases.
Each test case contains four integers:
n,
m,
z,
l
A1=0,
Ai=(Ai−1∗m+z)
mod
l
1≤m,z,l≤5∗105,
n=5∗105
Output
For every test.print the answer.
Sample Input
2 3 5 5 7 6 8 8 9
Sample Output
14 16
Source
2015 Multi-University Training Contest 5
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
int n,m,z,l;
int v[500015];
__int64 a[500015];
int b[500015];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d",&n,&m,&z,&l);
memset(v,0,sizeof(v));
v[0] = 1;
a[0] = 1;
int ph,pi;
int i;
int flag = 0;
m = m%l;
z = z%l;
for(i=2;i<=n;i++)
{
a[i] = ((a[i-1]*m)+z)%l;
//printf("a[%d] = %d\n",i,a[i]);
v[a[i]]++;
}
int k = 1;
for(int i=1;i<=n;i++)
{
if(v[a[i]]%2 == 1)
{
b[k] = a[i];
v[a[i]] = 0;
k++;
}
}
__int64 sum = 0;
__int64 sum1 = 0;
for(int i=1;i<k;i++)
{
sum = b[i] + b[i];
sum1^=sum;
}
printf("%I64d\n",sum1);
}
return 0;
}